sign in sign up
<<
Home , Augmented matrix, Band matrix, Bidiagonal matrix, Block matrix, Identity matrix, Invertible matrix

2.5. Inverse Matrices 81 2.5 Inverse Matrices

Suppose A is a square . We look for an “inverse matrix” A1 of the same size, such that A1 times A equals I . Whatever A does, A1 undoes. Their product is the —which does nothing to a vector, so A1Ax D x. But A1 might not exist. What a matrix mostly does is to multiply a vector x. Multiplying Ax D b by A 1 gives A1Ax D A1b. This is x D A1b. The product A1A is like multiplying by a number and then dividing by that number. A number has an inverse if it is not zero— matrices are more complicated and more interesting. The matrix A1 is called “A inverse.”

DEFINITION The matrix A is invertible if there exists a matrix A1 such that

A1A D I and AA1 D I: (1)

Not all matrices have inverses. This is the first question we ask about a : Is A invertible? We don’t mean that we immediately calculate A1. In most problems we never compute it! Here are six “notes” about A1. Note 1 The inverse exists if and only if elimination produces n pivots (row exchanges are allowed). Elimination solves Ax D b without explicitly using the matrix A 1. Note 2 The matrix A cannot have two different inverses. Suppose BA D I and also AC D I . Then B D C , according to this “proof by parentheses”:

B.AC/ D .BA/C gives BI D IC or B D C: (2)

This shows that a left-inverse B (multiplying from the left) and a right-inverse C (multi- plying A from the right to give AC D I ) must be the same matrix. Note 3 If A is invertible, the one and only solution to Ax D b is x D A1b:

Multiply Ax D b by A1: Then x D A1Ax D A1b:

Note 4 (Important) Suppose there is a nonzero vector x such that Ax D 0. Then A cannot have an inverse. No matrix can bring 0 back to x.

If A is invertible, then Ax D 0 can only have the zero solution x D A10 D 0.

Note 5 A 2 by 2 matrix is invertible if and only if ad bc is not zero: ab1 1 d b 2 2 D : by Inverse: cd ad bc ca (3)

This number ad bc is the of A. A matrix is invertible if its determinant is not zero (Chapter 5). The test for n pivots is usually decided before the determinant appears. 82 Chapter 2. Solving Linear Equations

Note 6 A matrix has an inverse provided no diagonal entries are zero: 2 3 2 3 d1 1=d1 6 : 7 1 6 : 7 If A D 4 :: 5 then A D 4 :: 5 :

dn 1=dn A D 12 Example 1 The 2 by 2 matrix 12 is not invertible. It fails the test in Note 5, because ad bc equals 2 2 D 0. It fails the test in Note 3, because Ax D 0 when x D .2; 1/. It fails to have two pivots as required by Note 1. Elimination turns the second row of this matrix A into a zero row.

The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. The a D 3 b D3 1 1 a C b D 0 numbers and have inverses 3 and 3 . Their sum has no inverse. ab D9 1 1 But the product does have an inverse, which is 3 times 3 . For two matrices A and B, the situation is similar. It is hard to say much about the invertibility of A C B. But the product AB has an inverse, if and only if the two factors A and B are separately invertible (and the same size). The important point is that A1 and B1 come in reverse order:

If A and B are invertible then so is AB. The inverse of a product AB is

.AB/1 D B1A1: (4)

To see why the order is reversed, multiply AB times B 1A1. Inside that is BB 1 D I : Inverse of AB .AB/.B1A1/ D AIA1 D AA1 D I:

We moved parentheses to multiply BB 1 first. Similarly B 1A1 times AB equals I . This illustrates a basic rule of : Inverses come in reverse order. It is also common sense: If you put on socks and then shoes, the first to be taken off are the . The same reverse order applies to three or more matrices:

Reverse order .ABC /1 D C 1B1A1: (5) Example 2 Inverse of an elimination matrix.IfE subtracts 5 times row 1 from row 2, then E1 adds 5 times row 1 to row 2: 2 3 2 3 100 100 E D 45105 and E1 D 45105 : 001 001

Multiply EE1 to get the identity matrix I . Also multiply E 1E to get I . We are adding and subtracting the same 5 times row 1. Whether we add and then subtract (this is EE 1/ or subtract and then add (this is E 1E/, we are back at the start. 2.5. Inverse Matrices 83

For square matrices, an inverse on one side is automatically an inverse on the other side. If AB D I then automatically BA D I . In that case B is A1. This is very useful to know but we are not ready to prove it. Example 3 Suppose F subtracts 4 times row 2 from row 3, and F 1 adds it back: 2 3 2 3 100 100 F D 40105 and F 1 D 40105 : 0 41 041

Now multiply F by the matrix E in Example 2 to find FE. Also multiply E 1 times F 1 to find .FE/1. Notice the orders FE and E1F 1! 2 3 2 3 10 0 100 FE D 4 51 05 is inverted by E1F 1 D 4 5 105 : (6) 20 41 0 4 1

The result is beautiful and correct. The product FE contains “20” but its inverse doesn’t. E subtracts 5 times row 1 from row 2. Then F subtracts 4 times the new row 2 (changed by row 1) from row 3. In this order FE, row 3 feels an effect from row 1. In the order E1F 1, that effect does not happen. First F 1 adds 4 times row 2 to row 3. After that, E1 adds 5 times row 1 to row 2. There is no 20, because row 3 doesn’t change again. In this order E 1F 1,row3 feels no effect from row 1.

In elimination order F follows E. In reverse order E1 follows F 1. E 1F 1 is quick. The multipliers 5, 4 fall into place below the diagonal of 1’s.

This special E 1F 1 and E1F 1G1 will be useful in the next sec- tion. We will explain it again, more completely. In this section our job is A1, and we expect some serious work to compute it. Here is a way to organize that computation.

Calculating A1 by Gauss-Jordan Elimination I hinted that A1 might not be explicitly needed. The equation Ax D b is solved by x D A1b. But it is not necessary or efficient to compute A1 and multiply it times b. Elimination goes directly to x. Elimination is also the way to calculate A 1,aswenow show. The Gauss-Jordan idea is to solve AA1 D I , finding each column of A1. 1 A multiplies the first column of A (call that x1/ to give the first column of I (call that e1/. This is our equation Ax1 D e1 D .1;0;0/. There will be two more equations. 1 Each of the columns x1, x2, x3 of A is multiplied by A to produce a column of I : 1 1 3 columns of A AA D A x1 x2 x3 D e1 e2 e3 D I: (7)

To invert a 3 by 3 matrix A, we have to solve three systems of equations: Ax 1 D e1 and 1 Ax2 D e2 D .0;1;0/and Ax3 D e3 D .0;0;1/. Gauss-Jordan finds A this way. 84 Chapter 2. Solving Linear Equations

The Gauss-Jordan method computes A1 by solving all n equations together. Usually the “” ŒA b has one extra column b. Now we have three right sides e1; e2; e3 (when A is 3 by 3). They are the columns of I , so the augmented matrix is really the ŒA I . I take this chance to invert my favorite matrix K, with 2’s on the main diagonal and 1’s next to the 2’s: 2 3 2 1 01 0 0 K 6 7 Start Gauss-Jordan on K e1 e2 e3 D 4 1 2 10 1 05 0 1 20 0 1 2 3 2 10100 ! 4 3 1 5 . 1 C / 0 2 1 2 10 2 row 1 row 2 0 12001 2 3 2 10100 4 0 3 1 1 105 ! 2 2 4 1 2 . 2 C / 003 3 3 1 3 row 2 row 3 We are halfway to K1. The matrix in the first three columns is U (upper triangular). The 2; 3 ; 4 pivots 2 3 are on its diagonal. Gauss would finish by back substitution. The contribution of Jordan is to continue with elimination! He goes all the way to the “reduced echelon form”. Rows are added to rows above them, to produce zeros above the pivots: 2 3 2 10100 Zero above ! 4 0 3 0 3 3 3 5 . 3 row 3 C row 2/ third pivot 2 4 2 4 4 004 1 2 1 2 3 3 3 3 3 1 200 1 . 2 row 2 C row 1/ 6 2 2 7 3 Zero above ! 4 0 3 0 3 3 3 5 second pivot 2 4 2 4 004 1 2 1 3 3 3 The last Gauss-Jordan step is to divide each row by its pivot. The new pivots are 1. We have reached I in the first half of the matrix, because K is invertible. The three columns of K1 are in the second half of ŒI K1 : 2 3 1 003 1 1 (divide by 2/ 6 4 2 4 7 6 7 3 / 6 1 1 7 1 (divide by 2 6 0 1 0 1 7 D I x1 x2 x3 D I K : 4 2 2 5 4 / (divide by 3 00 1 1 3 1 4 2 4

Starting from the 3 by 6 matrix ŒK I , we ended with ŒI K1 . Here is the whole Gauss-Jordan process on one line for any A: Gauss-Jordan Multiply AI by A1 to get Œ IA1: 2.5. Inverse Matrices 85

The elimination steps create the inverse matrix while changing A to I . For large matrices, we probably don’t want A1 at all. But for small matrices, it can be very worthwhile to know the inverse. We add three observations about this particular K 1 because it is an important example. We introduce the words symmetric, tridiagonal, and determinant:

1. K is symmetric across its main diagonal. So is K 1.

2. K is tridiagonal (only three nonzero ). But K 1 is a dense matrix with no zeros. That is another reason we don’t often compute inverse matrices. The inverse of a is generally a dense matrix. 2. 3 /. 4 / D 4 K 3. The product of pivots is 2 3 . This number 4 is the determinant of . 2 3 321 1 K 1 involves by the determinant K 1 D 42425 . (8) 4 123

This is why an invertible matrix cannot have a zero determnant. A1 A D 23 Example 4 Find by Gauss-Jordan elimination starting from 47 . There are two row operations and then a division to put 1’s in the pivots: 2310 2310 A I D ! this is UL1 4701 0121 2073 107 3 ! ! 2 2 this is I A1 : 0121 0121

That A1 involves division by the determinant ad bc D 2 7 3 4 D 2. The code for X D inverse.A/ can use rref, the “row reduced echelon form” from Chapter 3:

I D eye .n/I % Define the n by n identity matrix R D rref .ŒA I /I % Eliminate on the augmented matrix ŒA I X D R. W ;nC 1 W n C n/ % Pick A1 from the last n columns of R

A must be invertible, or elimination cannot reduce it to I (in the left half of R). Gauss-Jordan shows why A1 is expensive. We must solve n equations for its n columns.

To solve Ax D b without A1, we deal with one column b to find one column x.

In defense of A1, we want to say that its cost is not n times the cost of one system Ax D b. Surprisingly, the cost for n columns is only multiplied by 3. This saving is because the n equations Axi D ei all involve the same matrix A. Working with the right sides is relatively cheap, because elimination only has to be done once on A. The complete A1 needs n3 elimination steps, where a single x needs n3=3. The next section calculates these costs. 86 Chapter 2. Solving Linear Equations

Singular versus Invertible

We come back to the central question. Which matrices have inverses? The start of this section proposed the pivot test: A1 exists exactly when A has a full set of n pivots. (Row exchanges are allowed.) Now we can prove that by Gauss-Jordan elimination:

1. With n pivots, elimination solves all the equations Ax i D ei . The columns xi go into A1. Then AA1 D I and A1 is at least a right-inverse.

2. Elimination is really a sequence of by E’s and P ’s and D 1:

Left-inverse .D1 E P E/A D I: (9)

D1 divides by the pivots. The matrices E produce zeros below and above the pivots. P will exchange rows if needed (see Section 2.7). The product matrix in equation (9) is evidently a left-inverse. With n pivots we have reached A1A D I . The right-inverse equals the left-inverse. That was Note 2 at the start of in this section. So a square matrix with a full set of pivots will always have a two-sided inverse. Reasoning in reverse will now show that A must have n pivots if AC D I . (Then we deduce that C is also a left-inverse and CA D I .) Here is one route to those conclusions:

1. If A doesn’t have n pivots, elimination will lead to a zero row. 2. Those elimination steps are taken by an invertible M . So a row of MA is zero. 3. If AC D I had been possible, then MAC D M . The zero row of MA, times C , gives a zero row of M itself. 4. An invertible matrix M can’t have a zero row! A must have n pivots if AC D I .

That argument took four steps, but the outcome is short and important.

Elimination gives a complete test for invertibility of a square matrix. A 1 exists (and Gauss-Jordan finds it) exactly when A has n pivots. The argument above shows more:

If AC D I then CA D I and C D A1

Example 5 If L is lower triangular with 1’s on the diagonal, so is L1.

A is invertible if and only if no diagonal entries are zero.

Here L has 1’s so L1 also has 1’s. Use the Gauss-Jordan method to construct L1. Start by subtracting multiples of pivot rows from rows below. Normally this gets us halfway to the inverse, but for L it gets us all the way. L1 appears on the right when I appears on the left. Notice how L1 contains 11, from 3 times 5 minus 4. 2.5. Inverse Matrices 87

2 3 100100 Gauss-Jordan 43100105 D LI on triangular L 451001 2 3 100100 .3 times row 1 from row 2/ ! 40103105 .4 times row 1 from row 3/ ! 051401 .then 5 times row 2 from row 3/ 2 3 100100 40103105 D I L1 : ! 00111 51

1 L goes to I by a product of elimination matrices E32E31E21. So that product is L . All pivots are 1’s (a full set). L1 is lower triangular, with the strange entry “11”. 1 1 1 That 11 does not appear to spoil 3; 4; 5 in the good order E21 E31 E32 D L.

REVIEW OF THE KEY IDEAS

1. The inverse matrix gives AA1 D I and A1A D I . 2. A is invertible if and only if it has n pivots (row exchanges allowed). 3. If Ax D 0 for a nonzero vector x, then A has no inverse. 4. The inverse of AB is the reverse product B1A1. And .ABC /1 D C 1B1A1. 1 1 5. The Gauss-Jordan method solves AA D I to find the n columns of A . The augmented matrix AI is row-reduced to IA1 .

WORKED EXAMPLES

2.5 A The inverse of a triangular difference matrix A is a triangular sum matrix S: 2 3 2 3 100100 100100 AI D 4 1100105 ! 4 0101105 0 11001 0 11001 2 3 100100 ! 4 0101105 D IA1 D I sum matrix : 001111

If I change a13 to 1, then all rows of A add to zero. The equation Ax D 0 will now have the nonzero solution x D .1;1;1/. A clear signal: This new A can’t be inverted. 88 Chapter 2. Solving Linear Equations

2.5 B Three of these matrices are invertible, and three are singular. Find the inverse when it exists. Give reasons for noninvertibility (zero determinant, too few pivots, nonzero solution to Ax D 0) for the other three. The matrices are in the order A; B; C; D; S; E: 2 3 2 3 100 111 43 43 66 66 4 1105 4 1105 86 87 60 66 111 111

Solution 2 3 100 1 7 3 1 06 B1 D C 1 D S 1 D 4 1105 84 6 6 4 36 0 11

A is not invertible because its determinant is 4 6 3 8 D 24 24 D 0. D is not invertible because there is only one pivot; the second row becomes zero when the first row is subtracted. E is not invertible because a combination of the columns (the second column minus the first column) is zero—in other words Ex D 0 has the solution x D .1; 1; 0/. Of course all three reasons for noninvertibility would apply to each of A; D; E.

2.5 C Apply the Gauss-Jordan method to invert this triangular “” L. You see Pascal’s triangle—adding each entry to the entry on its left gives the entry below. The entries of L are “binomial coefficients”. The next row would be 1; 4; 6; 4; 1. 2 3 1 000 6 11007 Triangular Pascal matrix L D 6 7 D abs(pascal (4,1)) 4 1210 5 1331

Solution Gauss-Jordan starts with ŒL I and produces zeros by subtracting row 1: 2 3 2 3 1 0001000 1000 1000 6 0001007 6 100 1007 ŒL I D 6 11 7 ! 6 0 1 7 : 4 1210 00105 4 0 2101 0105 13310001 0 3311 001

The next stage creates zeros below the second pivot, using multipliers 2 and 3. Then the last stage subtracts 3 times the new row 3 from the new row 4: 2 3 2 3 1000 1000 1000 1 000 6 010011007 6 0100 007 !6 7!6 11 7D ŒI L1: 4 0 0 10 1 2 105 4 0010 1 210 5 0 0 31 2 3 01 00011331 All the pivots were 1! So we didn’t need to divide rows by pivots to get I . The inverse matrix L1 looks like L itself, except odd-numbered diagonals have minus signs. The same pattern continues to n by n Pascal matrices, L1 has “alternating diagonals”. 2.5. Inverse Matrices 89 Problem Set 2.5

1 Find the inverses (directly or from the 2 by 2 formula) of A; B; C : 03 20 34 A D B D C D : 40 and 42 and 57

2 For these “permutation matrices” find P 1 by trial and error (with 1’s and 0’s): 2 3 2 3 001 010 P D 40105 and P D 40015 : 100 100

3 Solve for the first column .x; y/ and second column .t; z/ of A1: 10 20 x 1 10 20 t 0 D D : 20 50 y 0 and 20 50 z 1 12 AA1 D I 1 A1 4 Show that 36 is not invertible by trying to solve for column of : 12 x 1 A 1 A1 D For a different , could column of 36 y 0 be possible to find but not column 2?

5 Find an upper triangular U (not diagonal) with U 2 D I which gives U D U 1. A AB D AC B D C 6 (a) If is invertible and , prove quickly that . A D 11 AB D AC (b) If 11 , find two different matrices such that . 7 (Important) If A has row 1 C row 2 D row 3, show that A is not invertible: (a) Explain why Ax D .1;0;0/cannot have a solution. (b) Which right sides .b1;b2;b3/ might allow a solution to Ax D b? (c) What happens to row 3 in elimination? 8 If A has column 1 C column 2 D column 3, show that A is not invertible: (a) Find a nonzero solution x to Ax D 0. The matrix is 3 by 3. (b) Elimination keeps column 1 C column 2 D column 3. Explain why there is no third pivot. 9 Suppose A is invertible and you exchange its first two rows to reach B. Is the new matrix B invertible and how would you find B 1 from A1? 10 Find the inverses (in any legal way) of 2 3 2 3 0002 3200 600307 643007 A D 6 7 B D 6 7 : 404005 and 400655 5000 0076 90 Chapter 2. Solving Linear Equations

11 (a) Find invertible matrices A and B such that A C B is not invertible. (b) Find singular matrices A and B such that A C B is invertible. 12 If the product C D AB is invertible .A and B are square), then A itself is invertible. Find a formula for A1 that involves C 1 and B. 13 If the product M D ABC of three square matrices is invertible, then B is invertible. (So are A and C .) Find a formula for B1 that involves M 1 and A and C . 14 If you add row 1 of A to row 2 to get B, how do you find B 1 from A1? 10 B D A : Notice the order. The inverse of 11 is

15 Prove that a matrix with a column of zeros cannot have an inverse. ab d b ad ¤ bc 16 Multiply cd times ca. What is the inverse of each matrix if ? 17 (a) What matrix E has the same effect as these three steps? Subtract row 1 from row 2, subtract row 1 from row 3, then subtract row 2 from row 3. (b) What single matrix L has the same effect as these three reverse steps? Add row 2 to row 3, add row 1 to row 3, then add row 1 to row 2. 18 If B is the inverse of A2, show that AB is the inverse of A. 19 Find the numbers a and b that give the inverse of 5 eye(4) – ones(4,4): 2 3 2 3 4 1 1 1 1 abbb 6141 17 6babb7 6 7 D 6 7 : 41 1415 4bbab5 1 1 14 bbba

What are a and b in the inverse of 6 eye(5) – ones(5,5)? 20 Show that A=4 eye(4) – ones(4,4) is not invertible: Multiply A ones(4,1). 21 There are sixteen 2 by 2 matrices whose entries are 1’s and 0’s. How many of them are invertible? Questions 22Ð28 are about the Gauss-Jordan method for calculating A 1. 22 Change I into A1 as you reduce A to I (by row operations): 1310 1410 AI D AI D 2701 and 3901

23 Follow the 3 by 3 text example but with plus signs in A. Eliminate above and below the pivots to reduce ŒA I to ŒI A1 : 2 3 210100 AI D 41210105 : 012001 2.5. Inverse Matrices 91

24 Use Gauss-Jordan elimination on ŒU I to find the upper triangular U 1: 2 3 2 3 2 3 1ab 100 1 4 5 4 5 4 5 UU D I 01c x1 x2 x3 D 010: 001 001

25 Find A1 and B1 (if they exist) by elimination on ŒA I and ŒB I : 2 3 2 3 211 2 1 1 A D 41215 and B D 41215 : 112 1 12

E E D1 A D 12 26 What three matrices 21 and 12 and reduce 26 to the identity matrix? 1 1 Multiply D E12E21 to find A .

27 Invert these matrices A by the Gauss-Jordan method starting with ŒA I : 2 3 2 3 100 111 A D 42135 and A D 41225 : 001 123

28 Exchange rows and continue with Gauss-Jordan to find A1: 0210 AI D : 2201

29 True or false (with a counterexample if false and a reason if true):

(a) A 4 by 4 matrix with a row of zeros is not invertible. (b) Every matrix with 1’s down the main diagonal is invertible. (c) If A is invertible then A1 and A2 are invertible.

30 For which three numbers c is this matrix not invertible, and why not? 2 3 2cc A D 4ccc5 : 87c

31 Prove that A is invertible if a ¤ 0 and a ¤ b (find the pivots or A1): 2 3 abb A D 4aab5 : aaa 92 Chapter 2. Solving Linear Equations

32 This matrix has a remarkable inverse. Find A1 by elimination on ŒA I . Extend toa5by5“alternating matrix” and guess its inverse; then multiply to confirm. 2 3 1 111 601117 A D 6 7 Ax D .1;1;1;1/: Invert 400115 and solve 0001

33 Suppose the matrices P and Q have the same rows as I but in any order. They are “permutation matrices”. Show that P Q is singular by solving .P Q/x D 0.

34 Find and check the inverses (assuming they exist) of these block matrices: I0 A0 0I : CI CD ID

35 Could a 4 by 4 matrix A be invertible if every row contains the numbers 0; 1; 2; 3 in some order? What if every row of B contains 0; 1; 2; 3 in some order?

36 In the Worked Example 2.5 C, the triangular Pascal matrix L has an inverse with “alternating diagonals”. Check that this L1 is DLD, where the D has alternating entries 1; 1; 1; 1. Then LDLD D I , so what is the inverse of LD D pascal (4,1)?

37 The Hilbert matrices have Hij D 1=.i C j 1/. Ask MATLAB for the exact 6 by 6 inverse invhilb(6). Then ask it to compute inv(hilb(6)). How can these be different, when the computer never makes mistakes?

38 (a) Use inv(P) to invert MATLAB’s 4 by 4 P D pascal(4). (b) Create Pascal’s lower triangular L D abs(pascal(4,1)) and test P D LLT.

39 If A D ones(4) and b D rand(4,1), how does MATLAB tell you that Ax D b has no solution? For the special b D ones(4,1), which solution to Ax Db is found by Anb?

Challenge Problems

40 (Recommended) A is a 4 by 4 matrix with 1’s on the diagonal and a; b; c on the diagonal above. Find A1 for this .

41 Suppose E1;E2;E3 are 4 by 4 identity matrices, except E1 has a; b; c in column 1 and E2 has d;e in column 2 and E3 has f in column 3 (below the 1’s). Multiply L D E1E2E3 to show that all these nonzeros are copied into L.

E1E2E3 is in the opposite order from elimination (because E3 is acting first). But E1E2E3 D L is in the correct order to invert elimination and recover A. 2.5. Inverse Matrices 93

42 Direct multiplications 1Ð4 give MM 1 D I , and I would recommend doing #3. M 1 shows the change in A1 (useful to know) when a matrix is subtracted from A: 1 M D I uv and M 1 D I C uv=.1 vu/. 1 change in I/ 2 M D A uv and M 1 D A1 C A1uvA1=.1 vA1u/ 1 1 3 M D I UV and M D In C U.Im VU/ V 4 M D A UW1V and M 1 D A1 C A1U.W VA1U/1VA1 The Woodbury-Morrison formula 4 is the “matrix inversion lemma” in engineering. The Kalman filter for solving block tridiagonal systems uses formula 4 at each step. The four matrices M 1 are in diagonal blocks when inverting these block matrices (v is 1 by n, u is n by 1, V is m by n, U is n by m). I u A u In U AU v 1 v 1 VIm VW

43 Second difference matrices have beautiful inverses if they start with T11 D 1 (instead of K11 D 2). Here is the 3 by 3 T and its inverse: 2 3 2 3 1 10 321 4 5 1 4 5 T 11 D 1 T D 121 T D 221 0 12 111

One approach is Gauss-Jordan elimination on ŒT I . That seems too mechanical. I would rather write T as the product of first differences L times U . The inverses of L and U in Worked Example 2.5 A are sum matrices, so here are T and T 1: 2 32 3 2 32 3 1 1 10 111 1 LU D 41154 1 15 U 1L1 D 4 115411 5 0 11 1 1 111 difference difference sum sum Question. (4by4) What are the pivots of T ? What is its 4 by 4 inverse? The reverse order UL gives what matrix T ? What is the inverse of T ?

44 Here are two more difference matrices, both important. But are they invertible? 2 3 2 3 2 101 1 100 612107 612107 C D 6 7 F D 6 7 : Cyclic 4 0 1215 Free ends 4 0 1215 1012 0011

One test is elimination—the fourth pivot fails. Another test is the determinant, we don’t want that. The best way is much faster, and independent of matrix size: Produce x ¤ 0 so that C x D 0. Do the same for F x D 0. Not invertible.

Show how both equations C x D b and F x D b lead to 0 D b1 C b2 CCbn. There is no solution for other b. 94 Chapter 2. Solving Linear Equations

45 Elimination for a 2 by 2 block matrix: When you multiply the first block row by CA1 and subtract from the second row, the “” S appears: I0AB AB A and D are square D CA1 I CD 0S S D D CA1B:

Multiply on the right to subtract A1B times block column 1 from block column 2. 2 3 233 AB I A1B AB D ‹ S D 44105 : 0S 0I Find for CI 401

The block pivots are A and S. If they are invertible, so is ŒA BI CD.

46 How does the identity A.I C BA/ D .I C AB/A connect the inverses of I C BA and I C AB? Those are both invertible or both singular: not obvious.